184_projects:transformers_sol

Back in Lakeview, the discovery of Dr. Watersa's illicit stash of jellybeans has shocked the town's residents. Not only has she been hiding delicious treats from her friends and neighbors, the town is also out-of-luck for the potential alternative energy source. In their anticipation of using the water mill for energy, the residents stupidly tore down all the ugly power lines. Now you have to help set them back up. You have access to a high-voltage power line from the edge of town, which has a voltage difference of $\Delta V = 240 \text{ kV } \cdot \sin(\omega t)$ compared to the ground, where $\omega = 60 \text{ Hz}$. You have access to the necessary safety equipment so you won't get electrocuted. You also have a plentiful supply of iron barring, which the town blacksmith has agreed to help you bend and mold as needed. Lastly, you have as much 12-gauge copper wire as you can imagine. Your goal is to devise a way to feed power lines to each house in Lakeview which don't exceed the maximum of 120 V compared to ground.

Power Line

Learning Goals

  • Explain how a voltage transformer works
  • Use the right hand rule to check the direction of the induced current
  • Explain what it means for a power line to have a “voltage”

Learning Issues

  • Creating the design of the transformer itself
  • Explaining how the magnetic field propagates in metal
  • Keeping all the variables straight by making sure everything is labelled

Alternate Solutions

  • Students might choose different designs for the ferromagnetic material as a part of their transformer

The main goal of this project is to design a step-down transformer. Ideally students should also be able to create a representation for how their step-down transformer can be placed into the power grid and provide 120-Volt power to residents of Lakeview.

Designing a step-up transformer is given in the notes, so this is something students should be able to look up, but for this project the difficulty lies in explaining how it works.

It makes sense intuitively to just flip the step-up transformer for our design. If the step-up transformer brings voltage up, we could just reverse it to bring the voltage down. Our design is shown below, where now the number of turns in the primary solenoid is much greater than the number of turns in the secondary solenoid – it's just the flipped step-up transformer.

Step-down Transformer

This representation shows only one way to create a metal object that you can wrap coils of wire around. Students may choose to create a different metal shape from the pipes, such as a horseshoe, or just a straight bar.

Guiding Questions to Motivate Transformer Design

  • Question: What is the general design of a transformer?
  • Expected Answer: Two solenoids wrapped around ferromagnetic material. Each solenoid is connected to ground on one end, and to a separate power line on the other end.
  • Question: What happens to ferromagnetic material when current exists in a solenoid wrapped around it?
  • Expected Answer: The material will magnetize! Electrons orbiting in the atoms that make up the material will align their movement with the magnetic field from the solenoid. The orbitals themselves will create their own magnetic field, and the material effectively becomes a “magnet”, whose magnetic field is aligned with the field from the solenoid.
  • Question: How do you know there will be current in the second solenoid?
  • Expected Answer: The magnetic field from the ferromagnetic material changes exactly how the magnetic field from the first solenoid changes. A changing magnetic field creates a changing magnetic flux in the second solenoid, which induces current/voltage.
  • Question: How can you control the induced current/voltage with your transformer design?
  • Expected Answer: The number of turns in each solenoid can be used to vary the magnetic flux in the solenoid. Changing magnetic flux is how we can figure out induced current/voltage.

As with the step-up transformer, the two solenoids are wrapped around the same iron ring. When current exists in the primary solenoid, it creates a magnetic field inside the solenoid, aligned along the iron ring. The magnetic field from the solenoid will cause all of the atoms within the iron to align with the magnetic field from the primary solenoid. Because iron atoms are very responsive to magnetic fields, even the atoms that are outside the primary solenoid will align with this magnetic field (largely because they are feeling the effects of their neighboring iron atoms). Because the magnetic field in the primary solenoid is oscillating (due to the alternating current), this means that the magnetic field in all of the iron ring is also constantly changing. For snapshot in time, the field may look like this:

Step-down Transformer with B-field

Discussion Questions

  • Question: Why do you need the iron between the coils?
  • Expected Answer: The iron magnetizes in the presence of the magnetic field from the primary solenoid. Magnetized bits of the iron influence the surrounding bits, until the whole piece of iron is magnetized. The iron is what produces the magnetic field inside the secondary solenoid. The iron is needed because the two solenoids cannot occupy the same space, so the primary magnetic field needs to be “reproduced” somehow at the location of the secondary solenoid.
  • Question: What would happen if the iron was removed? Would you still have an induced voltage?
  • Expected Answer: The primary solenoid produces a changing magnetic field, but it would not exist at the location secondary solenoid. So there would be no induced voltage, or at the most, very very little induced voltage from edge effects.
  • Question: What if you used a material that was not iron? Aluminum? PVC pipe?
  • Expected Answer: The magnetic field exists everywhere in the iron because it is a ferromagnetic material. Aluminum and PVC are not ferromagnetic, so they would not do the trick. It would be as if there was nothing there at all.

Important Approximation

  • How can you compare $B_\text{P}$ with $B_\text{S}$?

The iron is able to align its atoms with the magnetic field much faster than the current alternates between directions, which is why we draw the magnetic field the same everywhere. The iron also greatly amplifies the magnetic field that the primary solenoid would produce in air, so even though $B_\text{P}$ contains magnetic field contributions from the primary solenoid and from the iron, the contribution from the iron is far greater. For this reason, we approximate the magnetic field as the same at all locations in the iron. By this approximation, $B_\text{P} = B_\text{S}$.

Since the magnetic field in the iron ring is changing with the alternating current, (in the primary solenoid), this will induce a voltage ($V_\text{S}$) in the secondary solenoid. We can use Faraday's Law to write this as: $$-V_{\text{S}}=\frac{\text{d}\Phi_{B_{\text{S}}}}{\text{d}t}$$

The rest of our calculation follows just as it would for the analysis of the step-up transformer. We can rewrite the flux $\Phi_{B_{\text{S}}}$ using the flux definition: $$\Phi_{B_{\text{S}}}=\int \vec{B}_\text{S} \bullet \text{d}\vec{A}_\text{S}$$ where the area here would be the cross-sectional area of the iron ring (since this is where the magnetic field is). The direction of the magnetic field would always be perpendicular to the area of the cylinder - so the flux would simplfy to: $$\Phi_{B_{\text{S}}}=B_\text{S} A_\text{S}$$

But this is the magnetic flux through one loop in the secondary solenoid - if we want the total flux through all the loops then we have to multiply by the number of loops: $$\Phi_{B_{\text{S}}}=B_\text{S} A_\text{S} N_\text{S}$$

Evaluation Questions

  • Question: How did you simplify the dot product in the magnetic flux equation?
  • Expected Answer: The magnetic field in the ferromagnetic material is parallel with the area-vectors from the coils of the solenoid. In other words, the wire is not coiled around the barring at an angle.
  • Question: How did you simplify the integral in the magnetic flux equation?
  • Expected Answer: The magnetic field in the ferromagnetic material is constant in space. (Not constant in time!) This just means the magnetic field at the core of the iron barring is the same as near the edge. Every point inside the secondary solenoid has the same magnetic field, so $B$ can be pulled out of the integral. The remaining integral is just the cross-sectional area of the barring.
  • Question: Where does $N$ come from?
  • Expected Answer: The magnetic flux exists in each coil of the solenoid. We need to account for each coil, so it's as if the area were multiplied by $N$. Or we car just multiply the magnetic flux itself by $N$.

We can then plug this into the voltage equation we wrote above: $$-V_{\text{S}}=\frac{\text{d}}{\text{d}t}\biggl(B_\text{S} A_\text{S} N_\text{S}\biggr)$$

Since $N_\text{S}$ and $A_\text{S}$ are constant with respect to time (not adding/taking away loops or increasing/decreasing the area), these terms can come out of the derivative: $$-V_{\text{S}}=N_\text{S} A_\text{S} \frac{\text{d}}{\text{d}t}\biggl(B_\text{S}\biggr)$$

Important Assumption

  • How can you compare $A_\text{P}$ with $A_\text{S}$?

If we assume that the iron ring has a constant cross-sectional area, then $A_\text{P}=A_\text{S}$ and we already said that $B_\text{P}=B_\text{S}$. This means we can rewrite the flux through the secondary solenoid in terms of the magnetic field and area of the primary solenoid.

Using the assumption above, we can write: $$-V_{\text{S}}=N_\text{S} A_\text{P} \frac{\text{d}}{\text{d}t}\biggl(B_\text{P}\biggr)$$

If we look at the $A_\text{P}\frac{\text{d}}{\text{d}t}\biggl(B_\text{P}\biggr)$ term, this looks very much like the changing magnetic flux (or induced voltage) through a single loop of the primary solenoid: $$\frac{\text{d}\Phi_{B_{\text{P}}}}{\text{d}t}=\frac{\text{d}}{\text{d}t}\biggl(B_\text{P} A_\text{P}\biggr)$$

Discussion Question

  • Question: How can we relate the changing magnetic flux in the primary solenoid to the magnetic flux in the secondary solenoid?
  • Expected Answer: Write out each equation for changing magnetic flux. Using the assumptions above, the flux per coil can be set equal to each other. So then the changing magnetic fluxes will be proportional to each other based on the number of coils.

If we use Faraday's Law, we can rewrite the flux through a single loop of the primary solenoid in terms of the number of coils and the induced voltage in the whole solenoid: $$-V_\text{P}=N_\text{P} \frac{\text{d}}{\text{d}t}\biggl(B_\text{P} A_\text{P}\biggr)$$ $$\frac{\text{d}}{\text{d}t}\biggl(B_\text{P} A_\text{P}\biggr)=\frac{-V_\text{P}}{N_\text{P}}$$

Finally, We can plug this result into the $V_S$ equation above to get the potential in the secondary solenoid as it relates to the potential in the primary solenoid: $$-V_\text{S} = N_\text{S} \frac{-V_\text{P}}{N_\text{P}}$$ $$V_\text{S}=V_\text{P} \frac{N_\text{S}}{N_\text{P}}$$

Remember, we need to step down from the $240 \text{ kV}$ power line to a $120 \text{ V}$ line. This is a factor of 2000. One way to achieve this would be to set $N_\text{P} = 20000$, and $N_\text{S} = 10$. Due to the huge step down, it may be even easier to design a series of step-down transformers, so that we don't have to have such a large number of turns for the secondary solenoid.

Evaluation Questions

  • Question: How did you use the solenoid magnetic field equation?
  • Expected Answer: You don't actually need this equation. The only thing that matters is that each coil has the same area.
  • Question: How did you use the frequency of the alternating current?
  • Expected Answer: The frequency doesn't impact any of the calculations. The only way it plays into how the transformer works is that each line's voltage will have the same oscillation frequency.

Discussion Prompts

  • Question: Why does each solenoid need to be connected to ground?
  • Expected Answer: For the primary solenoid, that's how we get a high voltage across the solenoid. For the secondary solenoid, that's how we set the voltage of the power line (voltage from the power line to ground).

Extension Questions

  • Question: $N_\text{P}/N_\text{S}=2000$ is a pretty big ratio. How can you change your model so you won't need that big of a turn-ratio in a single transformer?
  • Expected Answer: Make two transformers, with ratios whose product is 2000. Maybe apply a factor of $N_\text{P}/N_\text{S}=40$ for one transformer, and then $N_\text{P}/N_\text{S}=50$ for a second transformer. This would be physically equivalent to just one step-down transformer with $N_\text{P}/N_\text{S}=2000$. See below for a possible design.

Two Step-down Transformers

  • 184_projects/transformers_sol.txt
  • Last modified: 2021/08/19 19:31
  • by dmcpadden